JEE Main 27-Aug-2021 Shift 2 Solved Paper

In, case I given, incident wavelength,
λ1=670.5nm
Stopping potential, V1=0.48V
In case II Incident wavelength,
λ2=474.6nm
Stopping potential is V2 .
By using Einstein photoelectric
equation, eV1=‌

hc

λ1

−φ0
where, h is Planck's constant
=6.63×10−34Js
c is speed of light =3×108ms−1
φ0 is work function
∴‌‌eV1=‌

hc

λ1

−φ0
‌ and similarly, ‌eV2=‌

hc

λ2

−φ0
⇒‌‌φ0=‌

hc

λ2

−eV2
From Eqs. (i) and (ii), we get
∴‌‌‌

hc

λ1

−eV1=‌

hc

λ2

−eV2
⇒eV1−eV2=‌‌hc(‌

1

λ1

−‌

1

λ2

)
⇒‌‌V1−V2=‌‌‌

hc

e

(‌

1

λ1

−‌

1

λ2

)
⇒‌‌‌‌V2=‌‌V1+‌

hc

e

(‌

1

λ2

−‌

1

λ1

)
=‌‌0.48+‌

6.63×10−34×3×108

1.6×10−19

=‌‌(‌

1

474.6

−‌

1

670.5

)
‌‌=0.48+12.4×10−7+9 (2.11×10−3−1.49×10−3)
=‌‌0.48+12.4×102−3×0.62
=‌‌0.48+0.77=1.25V